Решите неравенство 1)2√ х^2-7 х+12>1 2)√ х-1<3

Решите задачу:

$1)\; \; \sqrt{x^2-7x+12}\ \textgreater \ 1\; ,\\\\ODZ:\; x^2-7x+12 \geq 0,\\\\ +++(3)---(4)+++\; \; ,\; \; x\in (-\infty,3\, ]U[\, 4,+\infty)\\\\x^2-7x+12\ \textgreater \ 1\\\\x^2-7x+11\ \textgreater \ 0\\\\D=49-44=5\\\\x_1=\frac{7-\sqrt5}{2}\ \textless \ 3\; ,\; \; x_2=\frac{7+\sqrt5}{2}\ \textgreater \ 4\\\\x\in (\frac{7-\sqrt5}{2},3\, ]U[\, 4,\frac{7+\sqrt5}{2})$

$2)\; \; \sqrt{x-1}\ \textless \ 3\; ,\; ODZ:\; x-1 \geq 0\; ,\; \; x \geq 1\\\\x-1\ \textless \ 9\\\\x\ \textless \ 10\\\\Otvet:\; x\in [\, 1,10)$